What is the integral from 0 to 3 of (sqrt(x+1)) ?

The integral from 0 to 3 of (sqrt(x+1)) is 14/3

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integral from 0 to 3 of (sqrt(x+1))

\int_{0}^{3} (x + 1) d x

\frac{14}{3}

Decimal Notation

4.66666 \dots

Solution steps

\int_{0}^{3} x + 1 d x

Apply u-substitution

: \int_{1}^{4} u d u

= \int_{1}^{4} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{4}

= [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{4}

Compute the boundaries:

\frac{14}{3}

= \frac{14}{3}

f (x) = x^{2} - 6 x

\frac{d}{d x} (\frac{1}{2} arctan (\frac{x}{2}))

\frac{\partial}{\partial x} (\frac{( x - y )}{( z - y )})

y = \frac{x ^{2} + 32 x}{8}

\int \frac{x ^{n}}{n ^{2}} d x

What is the integral from 0 to 3 of (sqrt(x+1)) ?
The integral from 0 to 3 of (sqrt(x+1)) is 14/3